Geometry question: how to find the segment lengths of circumscribed triangle?

Question

Geometry question: how to find the segment lengths of circumscribed triangle?

Answer 1

ok i'm going to respond to Geometry concern enable one tangent element be X and the different be Y. enable the midsection of the circle be C. attitude XPC = attitude YPC so as that they are the two 30 levels. Triangle XPC is a suited triangle with a suited attitude at X and a 30 degree attitude at P the different area from the attitude, XC, is length 12 The sine of 30 levels is a million/2 so 12 is a million/2 of the hypotenuse and the hypotenuse is 24 The tangent section is the adjoining area so its length is sqrt(24^2 - 12^2) = 6 sqrt(3). Now i'm very upset you have asked 2 questions as a replace of one, yet I could desire to respond to this one too, or you're able to desire to finally finally end up giving suitable answer to somebody else as a replace. the two facets is sixteen meters. Use the midpoint to make a suited triangle with long leg 8. the attitude on the midsection is a million/6 of a circle that's 60 levels. If M is the midpoint, C is the midsection, and A is on the circle, the triangle sounds like C | M-------------A and attitude MCA is 60 levels, and MA is 8. radius = CA CA sin 60 = MA CA = 8 / (sqrt(3)/2) = sixteen/sqrt(3) = 9.237 approx.

Answer 2

ok i visit respond to Geometry situation permit one tangent factor be X and the different be Y. permit the middle of the circle be C. attitude XPC = attitude YPC so as that they are the two 30 stages. Triangle XPC is a amazing triangle with an excellent attitude at X and a 30 degree attitude at P the alternative ingredient from the attitude, XC, is length 12 The sine of 30 stages is a million/2 so 12 is a million/2 of the hypotenuse and the hypotenuse is 24 The tangent phase is the adjoining ingredient so its length is sqrt(24^2 - 12^2) = 6 sqrt(3). Now i'm very upset you have asked 2 questions particularly of one, yet I might desire to respond to this one too, or you should finally finally end up giving right answer to somebody else particularly. the two facets is sixteen meters. Use the midpoint to make an excellent triangle with long leg 8. the attitude on the middle is a million/6 of a circle that's 60 stages. If M is the midpoint, C is the middle, and A is on the circle, the triangle sounds like C | M-------------A and attitude MCA is 60 stages, and MA is 8. radius = CA CA sin 60 = MA CA = 8 / (sqrt(3)/2) = sixteen/sqrt(3) = 9.237 approx.

Answer 3

The two tangent segments from each vertex are congruent, so AP = AR, BP = BQ, and CQ = CR



So call them x, x, y, y, z, z and write three equations. Say the triangle lengths were

AB = 10, BC =12, CA =14



x + y = 10

y + z = 12

z + x = 14



Multiply row 1 by -1 and add to row 2 to get

z - x = 2



add that to row 3

2z = 16 so z = 8 and use that to get the other pieces.